JEE Mains Physics · Kinetic Theory

Molecular Speeds and Mean Free Path

Molecular speeds grow as √(T/M): the rms, average and most probable speeds are fixed multiples of √(RT/M), and the mean free path, the average distance between collisions, is 1/(√2 π d² n).

All Kinetic Theory notes JEE Mains Physics strategy

Why this matters

Thirty PYQs, all multiple choice, and three from 2026. Fourteen scale the rms speed with temperature or molar mass, six compare the three kinds of molecular speed or work one out from a molecule's mass, and ten use the mean free path or the collision frequency. Every ratio here needs kelvin temperatures, and every one has a square root that is easy to drop.

Concept 1 of 3: The rms speed scales as √(T/M)

Since ½m⟨v²⟩ = (3/2)kT, the rms speed is √(3kT/m) = √(3RT/M). Heat a gas to four times its kelvin temperature and its molecules go twice as fast; make them sixteen times heavier and they go four times slower. Two gases have equal rms speeds exactly when T/M is the same for both.

Definition

$v_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3kT}{m}} = \sqrt{\dfrac{3P}{\rho}}$ , with M in kg/mol.
Ratio form: $\dfrac{v_2}{v_1} = \sqrt{\dfrac{T_2}{T_1}\cdot\dfrac{M_1}{M_2}}$ .
Same T: the lighter gas is faster. Same $v_{rms}$ : $T/M$ is the same.
A diatomic molecule that splits into atoms halves M, which has the same effect on $v_{rms}$ as doubling T.
Vapour density is proportional to M, so at one temperature $v_{rms} \propto 1/\sqrt{\text{vapour density}}$ .
$\overline{v^{2}} = \dfrac{3RT}{M}$ is proportional to T: plotted against kelvin temperature it is a straight line through the origin.

rms speed

v_{rms} = \sqrt{\frac{3RT}{M}} \qquad \frac{v_2}{v_1} = \sqrt{\frac{T_2M_1}{T_1M_2}}

Worked example

Nitrogen (M = 28 g/mol) is at 77 °C. At what temperature does helium (M = 4 g/mol) have the same rms speed? Find that speed.

(R = 8.31)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 2 · Q5Moderate

Example 1 · Kinetic Theory · Molecular Speeds and Mean Free Path

The r.m.s speed of oxygen molecules at

47^{\circ}C

is equal to that of the hydrogen molecules kept at

\_\_\_\_

\ ^{\circ}C

. (Mass of oxygen molecule/mass of hydrogen molecule=32/2)

Speed goes as √T, not T

Four times the kelvin temperature gives twice the speed, not four times. The same square root applies to the molar mass.

Celsius in the ratio

Equal T/M must use kelvin. With nitrogen at 77 °C, 77/28 would put helium at 11 °C instead of the true 50 K.

Drill 13 more on the rms speed scales as √(t/m)

Concept 2 of 3: The rms, average and most probable speeds

Molecular speeds are spread out (the Maxwell distribution), so there are three useful averages. The most probable speed sits at the peak of the curve, the average speed is a little higher, and the rms speed, which weights fast molecules more because it squares the speeds, is higher still. All three are fixed multiples of √(RT/M), so their ratios are the same for every gas at every temperature.

Definition

Most probable: $v_p = \sqrt{\dfrac{2RT}{M}} \approx 1.41\sqrt{\dfrac{RT}{M}}$ .
Average: $\bar{v} = \sqrt{\dfrac{8RT}{\pi M}} \approx 1.60\sqrt{\dfrac{RT}{M}}$ .
Root mean square: $v_{rms} = \sqrt{\dfrac{3RT}{M}} \approx 1.73\sqrt{\dfrac{RT}{M}}$ .
$v_p : \bar{v} : v_{rms} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3}$ . Divide any two; T and M cancel.
With the mass of one molecule: $v_{rms} = \sqrt{\dfrac{3kT}{m}}$ .
Brownian particles also carry $\tfrac{3}{2}kT$ on average, so $v_{rms} = \sqrt{3kT/m}$ holds for a smoke particle too; it is tiny because m is huge.

Three molecular speeds

v_p = \sqrt{\frac{2RT}{M}} \qquad \bar{v} = \sqrt{\frac{8RT}{\pi M}} \qquad v_{rms} = \sqrt{\frac{3RT}{M}}

Worked example

Find the most probable, average and rms speeds of helium molecules (M = 4 g/mol) at 400 K.

(R = 8.3)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · JEE Mains 2022 — 25 June · Q7Moderate

Example 2 · Kinetic Theory · Molecular Speeds and Mean Free Path

The relation between root mean square speed

\left( v_{rms} \right)

and most probable speed

\left( v_{p} \right)

for the molar mass

M

of oxygen gas molecule at the temperature of

300\text{ }K

will be :-

Using the rms speed when the average speed is asked

"Average speed" is √(8RT/πM); "rms speed" is √(3RT/M). They differ by about 8%, enough to land on a wrong option.

Molar mass in grams

With R = 8.31 J mol⁻¹ K⁻¹, M must be in kg/mol: 32 g/mol is 0.032 kg/mol. Grams make every speed about 32 times too small.

Drill 5 more on the rms, average and most probable speeds

Concept 3 of 3: Mean free path and collision frequency

Picture a molecule of diameter d moving along: it hits every molecule whose centre comes within d of its path, so it sweeps out a tube of cross-section πd². The more molecules per cubic metre and the bigger they are, the sooner it hits one. Allowing for the motion of the others adds a factor √2. Divide the average speed by the mean free path and you get the number of collisions per second.

Definition

$\lambda = \dfrac{1}{\sqrt{2}\,\pi d^{2}n}$ , n = molecules per m³. With $P = nkT$ : $\lambda = \dfrac{kT}{\sqrt{2}\,\pi d^{2}P}$ .
$\lambda \propto \dfrac{1}{d^{2}}$ and $\propto \dfrac{1}{n}$ . At constant pressure $\lambda \propto T$ ; in a sealed rigid vessel n is fixed, so $\lambda$ does not change on heating.
Collision frequency $Z = \dfrac{\bar{v}}{\lambda} = \sqrt{2}\,\pi d^{2}n\bar{v}$ ; mean time between collisions $\tau = 1/Z$ .
Sealed vessel heated: n fixed, so $Z \propto \bar{v} \propto \sqrt{T}$ .
Two gases at the same T and n: $Z \propto \dfrac{d^{2}}{\sqrt{m}}$ .

Mean free path

\lambda = \frac{1}{\sqrt{2}\,\pi d^{2}n} = \frac{kT}{\sqrt{2}\,\pi d^{2}P} \qquad Z = \frac{\bar{v}}{\lambda}

Worked example

Molecules of diameter

2 \times 10^{-10}\ \text{m}

are in a gas at 300 K and

1.0 \times 10^{5}\ \text{Pa}

, with an average speed of 700 m/s. Find the mean free path and the collision frequency.

(k = 1.38 \times 10^{-23}\ \text{J/K})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q4Moderate

Example 3 · Kinetic Theory · Molecular Speeds and Mean Free Path

The mean free path of a molecule of diameter

5 \times 10^{- 10}\text{ }m

at the temperature

41^{\circ}C

and pressure

1.38 \times 10^{5}\text{ }Pa

, is given as

\_\_\_\_

m. (Given

k_{B} = 1.38 \times 10^{- 23}\text{ }J/K

1/d instead of 1/d²

The mean free path goes as 1/d²: the target is an area, πd². Halving the diameter makes the path four times longer, not twice.

Heating at constant pressure versus in a sealed vessel

At constant pressure the gas spreads out and λ grows with T. In a sealed rigid vessel the molecules per cubic metre do not change, so λ stays the same and only the speed, and with it Z, goes up.

Drill 9 more on mean free path and collision frequency

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The rms speed scales as √(T/M)
rms speed
$v_{rms} = \sqrt{\frac{3RT}{M}} \qquad \frac{v_2}{v_1} = \sqrt{\frac{T_2M_1}{T_1M_2}}$
The rms, average and most probable speeds
Three molecular speeds
$v_p = \sqrt{\frac{2RT}{M}} \qquad \bar{v} = \sqrt{\frac{8RT}{\pi M}} \qquad v_{rms} = \sqrt{\frac{3RT}{M}}$
Mean free path and collision frequency
Mean free path
$\lambda = \frac{1}{\sqrt{2}\,\pi d^{2}n} = \frac{kT}{\sqrt{2}\,\pi d^{2}P} \qquad Z = \frac{\bar{v}}{\lambda}$

Watch out for (6)

Speed goes as √T, not T
→ The rms speed scales as √(T/M)
Celsius in the ratio
→ The rms speed scales as √(T/M)
Using the rms speed when the average speed is asked
→ The rms, average and most probable speeds
Molar mass in grams
→ The rms, average and most probable speeds
1/d instead of 1/d²
→ Mean free path and collision frequency
Heating at constant pressure versus in a sealed vessel
→ Mean free path and collision frequency

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